Solve $ \left|\frac{x}{x+2}\right|\leq 2 $

Question

I am experiencing a little confusion in answering a problem on Absolute Value inequalities which I just started learning. This is the problem: Solve: $$ \left|\frac{x}{x+2}\right|\leq 2 $$ The answer is given to be $x\leq-4$ or $x \geq-1$

This is my attempt to solve the problem:

By dividing, $\left|\frac{x}{x+2}\right|\leq 2$ is equivalent to $\left |1-\frac{2}{x+2}\right|\leq2$ which is also equivalent to $\left |\frac{2}{x+2}-1\right|\leq2$

So, $-2\leq\frac{2}{x+2}-1\leq2$ which is equivalent to $-\frac{1}{2}\leq\frac{1}{x+2}\leq\frac{3}{2}$

Case 1: $x+2>0$. Solving $-\frac{1}{2}\leq\frac{1}{x+2}\leq\frac{3}{2}$, I get $x\geq-4$ and $x\geq-\frac{4}{3}$ which is essentially $x\geq-\frac{4}{3}$.

Case 2: $x+2<0$. Solving $-\frac{1}{2}\times(x+2)\geq{1}\geq\frac{3}{2}\times(x+2)$, I get $x\leq-4$ and $x\leq-\frac{4}{3}$ which is essentially $x\leq-4$.

So, the solutions are: $x\leq-4$ or $x\geq-\frac{4}{3}$.

I couldn't get $x \geq-1$ as a solution. Did I do anything wrong? The book I am using is Schaum's Outlines-Calculus. Another question I would like to ask is that am I using 'and' and 'or' correctly in the above attempt to solve the problem? I have had this problem many times.

Answer 1

So you have $$ \lvert x \rvert \leq 2\lvert x+2\rvert. $$ If $x \geq 0$, then $x + 2 > 0$. And then the equation reads $x \leq 2x + 4$ implying that $x \geq -4 $. This in all gives you the solutions $x \geq 0$.

If $x\in [-2, 0)$, then the equation becomes $-x \leq 2x + 4$. So $-4\leq 3x$. So $\frac{-4}{3} \leq x$. In all you get $x\in [-4/3, 0)$.

Now the last case is where $x <-2$. Then $-x \leq -2x - 4$. So $4 \leq -x$. So $-4\geq x$. This gives the set of solutions $(-\infty, -4]$.

Putting it all together gives you the closed interval: $$ (-\infty, -4] \cup [-4/3,\infty) $$

It looks like your solution is correct.

Answer 2

the case $x+2>0$, you do not need to solve $-1/2\le 1/(x+2)\leftarrow x+2>0$.

the case $x+2<0$, the inequality you solve it's not the original one. It should be $-1/2 \le 1/(x+2) \le 3/2$ and in case $x+2<0\rightarrow 1/(x+2)<3/2$, so it's just $-1/2 \le 1/(x+2)$.

Think it simply.

Answer 3

To solve this kind of problem systematically. If you want to solve $|f(x)|\le c$ or $|f(x)|\geq c$ (for some constant $c\geq 0$), you have to study the sign of $f(x)$. It can be longer for "simple" problem like yours, but it has the advantage to always give you the solution for more complex ones.

For your particular example, we have $f(x)=\frac{x}{x+2}$, which is: $$\begin{array}{c|ccccc} & &-2 & &0 & &\\ \hline x & - &- & - &0 &+\\ x+2 & - & 0 & + & + & +\\ \hline f(x) & + &\nexists &- & 0 &+ \end{array}$$ Which means that $f(x)\geq 0$ when $x\in(-\infty,-2)\cup[0,+\infty)$ and $f(x)\le 0$ when $x\in(-2,0]$.

1) For $x\in(-\infty,-2)\cup[0,+\infty)$, $f(x)\geq 0$, so that $$\left\vert\frac{x}{x+2}\right\vert=\frac{x}{x+2}$$ and the inequation is $$\frac{x}{x+2}\le 2\iff\frac{-x-4}{x+2}\le 0$$ which gives: $$\begin{array}{c|ccccc} & &-4 & &-2 & &\\ \hline -x-4 & + &0 & - &- &-\\ x+2 & - & - & - & 0 & +\\ \hline & - &0 &+ & \nexists &- \end{array}$$ and the solutions are $$\left[(-\infty,-4]\cup(-2,+\infty)\right]\cap\left[(-\infty,-2)\cup[0,+\infty)\right]=(-\infty,-4]\cup[0,+\infty)$$

2) For $x\in]-2,0]$, $f(x)\le 0$, so that $$\left\vert\frac{x}{x+2}\right\vert=-\frac{x}{x+2}$$ and the inequation is $$\frac{x}{x+2}\le 2\iff\frac{-3x-4}{x+2}\le 0$$ which gives: $$\begin{array}{c|ccccc} & &-2 & &-4/3 & &\\ \hline -3x-4 & + &+ & + &0 &-\\ x+2 & - & 0 & + & + & +\\ \hline & - &\nexists &+ & 0 &- \end{array}$$ and the solutions are $$\left[(-\infty,-2)\cup[-4/3,+\infty)\right]\cap(-2,0]=[-4/3,0]$$

And the final solution is given by $$\left\vert\frac{x}{x+2}\right\vert\le 2\iff x\in(-\infty,-4]\cup[0,+\infty)\cup[-4/3,0]=(-\infty,-4]\cup[-4/3,+\infty)$$

I know it seems long in this particular case but it is good to have a general method for this kind of problem.

Answer 4

Given that, $$\left|\frac{x}{x+2}\right|\le 2$$$$\iff |x|\le 2|x+2| $$ Notice, $x=-2$ & $x=0$ are two critical points on the number line. Now, let's consider the following cases,

Case 1: If $\color{blue}{x\le -2}$ $$-x\le -2(x+2)\iff x\le -4\ \ \ ({\text{True}})$$ $$\implies \color{red}{x\in(-\infty, -4]}$$ Case 2: If $\color{blue}{-2<x<0}$ $$-x\le 2(x+2)\iff x\ge -\frac{4}{3}\ \ \ $$ but, $x<0$ hence, $$-\frac{4}{3}<x<0\iff \color{red}{x\in\left[-\frac{4}{3}, 0\right)}$$

Case 3: If $\color{blue}{x\ge 0}$ $$x\le 2(x+2)\iff x\ge -4\ \ \ $$ but, $x\ge 0$ hence, $$x\ge 0\iff \color{red}{x\in\left[0, \infty\right)}$$ Hence, the combining the above cases, the complete solution is given as $$\color{red}{x\in (-\infty, -4]\cup \left[-\frac{4}{3}, \infty\right) }$$

Answer 5

Too long for a comment:

As long as $x\in\Bbb{R}$, you can consider $|\frac{x}{x+2}|$ as $(\frac{x}{x+2})\text{sign}(\frac{x}{x+2})$